Find the equation of the line that is perpendicular to Y=6x+3 and passes through the point (-24,5)
1 answer:
The slope of the given line is the coefficient of x, 6. The slope of the perpendicular line will be the negative reciprocal of that, -1/6. Then the point-slope form of the equation of the line is
y = m(x -h) +k . . . . . . line with slope m through point (h, k)
You want m=-1/6, (h, k) = (-24, 5), so your equation is
y = (-1/6)(x +24) +5
y = (-1/6)x +1
y = m(x -h) +k . . . . . . line with slope m through point (h, k)
You want m=-1/6, (h, k) = (-24, 5), so your equation is
y = (-1/6)(x +24) +5
y = (-1/6)x +1
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Step-by-step explanation:
We will use pythagoras' Theorem for this question
where c is the longest side (in this case, the diagonal)
a and b are the 2nd and 3rd longest side (interchangeable)
given a = 10.6, b = 16.8,